如何在 JavaScript 数组上使用模和按位 XOR 运算符反转算法？

Question

var alphaNumbersStr = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789';
var hexaString      = '0123456789abcdef';
var str32           = '4af27d7ef70b1263da25022af735508b';
var str             = 'FoooooBar';

var arr = [];
for (var i = 0; i < str.length; i++) {
    arr[i] = alphaNumbersStr.indexOf(str.charAt(i));
}
// --
// Start : arr ==> [ 31, 14, 14, 14, 14, 14, 27, 0, 17 ]
// --
for (var i = 0; i < 10000; i++) {

    arr[i % str.length] += hexaString.indexOf(str32.charAt(i % 32)) ^ arr[(i + 1) % str.length] ^ arr[(i + 2) % str.length];

    if (arr[i % str.length] > alphaNumbersStr.length - 1) {
        arr[i % str.length] %= alphaNumbersStr.length;
    }
}
// --
// End : after 10000 iterations : arr ==> [ 38, 8, 32, 4, 18, 24, 2, 48, 22 ]
// --

鉴于以下代码片段，我正在尝试反转此算法，以便我可以 return 到原始数组状态（步骤 0）。

知道如何解决这个问题吗？

Answer 1

只需向后执行所有步骤。执行的操作使得模运算可以恢复。

function revert(arr) {                                                                
    for (let i = 9999; i >= 0; i--) {
        //we know this as these values didnt get affected this iteration
        let added = hexaString.indexOf(str32.charAt(i % 32)) ^ arr[(i + 1) % str.length] ^ arr[(i + 2) % str.length];

        //the important difference in modulo
        let modadd = added % alphaNumbersStr.length;

        //single overflow is what we need to predict
        //if we went over, we went smaller again
        let overflow = modadd > arr[i % str.length];

        //we need to undo the modulo
        if (overflow) {
            let diff = modadd - arr[i % str.length];
            arr[i % str.length] = alphaNumbersStr.length - diff;
        }
        //Not much to do otherwise
        else arr[i % str.length] -= modadd;
    }
    return arr;
}

//Just so it can be run
function change(){var result,value=document.getElementById("a").value;try{result=algorithm(eval(value))}catch(a){result=["error"]}document.getElementById("t").innerHTML="["+result.join(", ")+"]",document.getElementById("i").innerHTML="["+(result[0]==='error' ? result : revert(result)).join(", ")+"]"}function algorithm(a){for(var b=0;b<1e4;b++)a[b%str.length]+=hexaString.indexOf(str32.charAt(b%32))^a[(b+1)%str.length]^a[(b+2)%str.length],a[b%str.length]>alphaNumbersStr.length-1&&(a[b%str.length]%=alphaNumbersStr.length);return a}var alphaNumbersStr="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789",hexaString="0123456789abcdef",str32="4af27d7ef70b1263da25022af735508b",str="FoooooBar";change();

Value of <code>arr</code>:&nbsp;&nbsp;<input onblur="change()" id="a" size="50" value="[31, 14, 14, 14, 14, 14, 27, 0, 17]">
<hr>
Through algorithm:&nbsp;&nbsp;<code id="t"></code><br>
Through revert:&nbsp;&nbsp;<code id="i"></code>

如何在 JavaScript 数组上使用模和按位 XOR 运算符反转算法？

How to reverse an algorithm using modulo and bitwise XOR operators on a JavaScript array?

javascript

algorithm

hash

bitwise-operators

modulo